In January 2013, I wrote a post explaining my use of Beeminder, after using it for six months. Well, I’ve now been using it continuously for almost four years! It has become such an integral part of my life and workflow that I literally don’t know what I would do if it went away. So I decided it was high time to write another blog post about Beeminder. This time, instead of enumerating things I am currently using it for, I will focus on things I have accomplished with the help of Beeminder. There is little doubt in my mind that I am much awesomer today than I would have been without Beeminder.

First, what is Beeminder? Here’s what I wrote three and a half years ago, which I think is still a good description:

The basic idea is that it helps you keep track of progress on any quantifiable goals, and gives you short-term incentive to stay on track: if you don’t, Beeminder takes your money. But it’s not just about the fear of losing money. Shiny graphs tracking your progress coupled with helpfully concrete short-term goals (“today you need to write 1.3 pages of that paper”) make for excellent positive motivation, too.

The key property that makes Beeminder work so well for me is that it makes long-term goals into short-term ones. I am a terrible procrastinator—due to hyperbolic discounting I can be counted on to pretty much ignore anything with long-term rewards or consequences. A vague sense that I ought to take better care of my bike is not enough to compel me to action in the present; but “inflate your tires and grease your chain before midnight or else pay $5” is.

Over the past three years, I have spent about 1 hour per week (typically spread over 3 or 4 days) learning biblical Hebrew. It adds up to almost 150 hours of Hebrew study—which doesn’t sound like a whole lot, but almost every minute of it was quality, focused study time. And since it has been so spread out, the material is quite firmly embedded in my long-term memory. I recently finished working through the entire introductory textbook I was using, while doing every single exercise in the associated workbook. I am still far from being an expert, but I can actually read simple things now.

Since September I have been swimming two mornings a week: when I started, I could barely do two laps before feeling like I was going to be sick; now, I can swim 500m in under 9 minutes (just under double world record pace =).

There are lots of other things I use Beeminder for, but these are the accomplishments I am proudest of. If you want to do awesome things but can never quite seem to find the time or motivation to do them, give it a try!

In a previous post I defined the network reliability problem. Briefly, we are given a directed graph whose edges are labelled with probabilities, which we can think of as giving the likelihood of a message successfully traversing a link in a network. The problem is then to compute the probability that a message will successfully traverse the network from a given source node to a given target node.

Several commenters pointed out the connection to Bayesian networks. I think they are right, and the network reliability problem is a very special case of Bayesian inference. However, so far this hasn’t seemed to help very much, since the things I can find about algorithms for Bayesian inference are either too general (e.g. allowing arbitrary functions at nodes) or too specific (e.g. only working for certain kinds of trees). So I’m going to put aside Bayesian inference for now; perhaps later I can come back to it.

Star semirings and path independence

Consider the related problem of computing the reliability of the single most reliable path from to in a network. This is really just a disguised version of the shortest path problem, so one can solve it using Dijkstra’s algorithm. But I want to discuss a more general way to think about solving it, using the theory of star semirings. Recall that a semiring is a set with two associative binary operations, “addition” and “multiplication”, which is a commutative monoid under addition, a monoid under multiplication, and where multiplication distributes over addition and . A star semiring is a semiring with an additional operation satisfying . Intuitively, (though can still be well-defined even when this infinite sum is not; we can at least say that if the infinite sum is defined, they must be equal). If is a star semiring, then the semiring of matrices over is also a star semiring; for details see Dolan (2013), O’Connor (2011), Penaloza (2005), and Lehmann (1977). In particular, there is a very nice functional algorithm for computing , with time complexity (Dolan 2013). (Of course, this is slower than Dijkstra’s algorithm, but unlike Dijkstra’s algorithm it also works for finding shortest paths in the presence of negative edge weights—in which case it is essentially the Floyd-Warshall algorithm.)

Now, given a graph and labelling , define the adjacency matrix to be the matrix of edge probabilities, that is, . Let be the star semiring of probabilities under maximum and multiplication (where , since ). Then we can solve the single most reliable path problem by computing over this semiring, and finding the largest entry. If we want to find the actual most reliable path, and not just its reliability, we can instead work over the semiring , i.e. probabilities paired with paths. You might enjoy working out what the addition, multiplication, and star operations should be, or see O’Connor (2011).

In fact, as shown by O’Connor and Dolan, there are many algorithms that can be recast as computing the star of a matrix, for an appropriate choice of semiring: for example, (reflexive-)transitive closure; all-pairs shortest paths; Gaussian elimination; dataflow analysis; and solving certain knapsack problems. One might hope that there is similarly an appropriate semiring for the network reliability problem. But I have spent some time thinking about this and I do not know of one.

Consider again the simple example given at the start of the previous post:

For this example, we computed the reliability of the network to be , by computing the probability of the upper path, , and the lower path, , and then combining them as , the probability of success on either path less the double-counted probability of simultaneous success on both.

Inspired by this example, one thing we might try would be to define operations and . But when we go to check the semiring laws, we run into a problem: distributivity does not hold! , but . The problem is that the addition operation implicitly assumes that the events with probabilities and are independent: otherwise the probability that they both happen is not actually equal to . The events with probabilities and , however, are not independent. In graph terms, they represent two paths with a shared subpath. In fact, our example computation at the beginning of the post was only correct since the two paths from to were completely independent.

I’m in Memphis (again!) this weekend attending the 2016 CCSC Midsouth Regional Conference, which also has a student programming contest attached. I’m very proud of the two teams from Hendrix, who placed first and fifth out of 17 teams. This is now the third year in a row that a team from Hendrix has won the contest (though I had nothing to do with the first two!). The members of the winning team are all graduating this year, but each of the members of the other team will be around at least one more year, and I have talked to quite a few other students who are interested. I’m excited to continue building up a programming team with a tradition of excellence.

I’ve been meaning for a while to put the source files for my CIS 194 materials in a publically accessible place, and I’ve finally gotten around to it: you can now find everything in byorgey/haskell-course on github.

I make no particular guarantees about anything; e.g. there is a crufty, complicated shake script that builds everything, but it probably doesn’t even compile with the latest version of Shake.

There are some obvious next steps, for which I have not the time:

Get everything building again

Fix any outstanding typos, confusions, etc.

Update it to be more specifically for general-audience learning rather than for a course at Penn

Find a permanent web home for the material. It’s sort of a happy accident that it has been accessible on the Penn site for so long, but it’s not a stable situation; I don’t even know who would have access to that site anymore.

tl;dr: I know how to generate random instances of data types in a generic way, and even have some old code that already does all the hard work, but won’t have time to polish and package it until this summer. If you’re interested in helping, let me know!

This morning Kenny Foner pointed out to me this tweet by Gabriel Gonzales, asking why there isn’t a default Arbitrary instance for types implementing Generic. It reminded me that I’ve been meaning for a while now (years, in fact!) to get around to packaging up some code that does this.

As several pointed out on Twitter, this seems obvious, but it isn’t. It’s easy to write a generic Arbitrary instance, but hard to write one that generates a good distribution of values. The basic idea is clear: randomly pick a constructor, and then recursively generate random subtrees. The problem is that this is very likely to either blow up and generate gigantic (even infinite) trees, or to generate almost all tiny trees, or both. I wrote a post about this three years ago which illustrates the problem. It also explains half of the solution: generate random trees with a target size in mind, and throw out any which are not within some epsilon of the target size (crucially, stopping the generation early as soon as the tree being generated gets too big).

However, I never got around to explaining the other half of the solution: it’s crucially important to use the right probabilities when picking a constructor. With the wrong probabilities, you will spend too much time generating trees that are either too small or too big. The surprising thing is that with exactly the right probabilities, you can expect to wait only time before generating a tree of size (approximately1) .2

So, how does one pick the right probabilities? Essentially, you turn the generic description of your data type into a mutually recursive system of generating functions, and (numerically) find their radii of convergence, when thought of as functions in the complex plane. Using these values it is straightforward to compute the right probabilities to use. For the intrepid, this is explained in Duchon et. al3.

I have some old Haskell code from Alexis Darrasse which already does a bunch of the work. It would have to be updated a bit to work with modern libraries and with GHC.Generics, and packaged up to go on Hackage. I won’t really have time to work on this until the summer—but if anyone else is interested in working on this, let me know! I’d be happy to send you the code and provide some guidance in figuring it out.

The constant factor depends on how approximate you are willing to be.↩

I wanted to put an exclamation point at the end of that sentence, because this is really surprising. But it looked like factorial. So, here is the exclamation point: !↩

This weekend I’m in Memphis for SIGCSE 2016. This is my first time at SIGCSE, so I’m looking forward to picking up some new ideas in CS education, and more importantly to meeting lots of new people. If you’re here too, feel free to say hi!

Let be a directed graph with vertices and edges . Multiple edges between the same pair of vertices are allowed. For concreteness’ sake, think of the vertices as routers, and the edges as (one-way) connections. Let denote the set of probabilities, and be a function which assigns some probability to each edge. Think of as the probability that a single message sent along the edge from the source router will successfully reach the target router on the other end.

Suppose that when a router receives a message on an incoming connection, it immediately resends it on all outgoing connections. For , let denote the probability that, under this “flooding” scenario, at least one copy of a message originating at will eventually reach .

For example, consider the simple network shown below.

A message sent from along the upper route through has an probability of arriving at . By definition a message sent along the bottom route has an probability of arriving at . One way to think about computing the overall probability is to compute the probability that it is not the case that the message fails to traverse both links, that is, . Alternatively, in general we can see that , so as well. Intuitively, since the two events are not mutually exclusive, if we add them we are double-counting the situation where both links work, so we subtract the probability of both working.

The question is, given some graph and some specified nodes and , how can we efficiently compute ? For now I am calling this the “network reliability problem” (though I fully expect someone to point out that it already has a name). Note that it might make the problem a bit easier to restrict to directed acyclic graphs; but the problem is still well-defined even in the presence of cycles.

This problem turned out to be surprisingly more difficult and interesting than it first appeared. In a future post or two I will explain my solution, with a Haskell implementation. In the meantime, feel free to chime in with thoughts, questions, solutions, or pointers to the literature.